Java - Calculating Pi

Question

I've been wanting to calculate pi with the Gregory-Leibniz series, and there's something wrong with it. I'm a bit of a starter to Java, so this may not be a hard question to answer. I don't know what's going on, so can somebody help me please?

Code so far:

package com.waitdev.pi;

public class Pi {

    static int pi = 0;

    public static void main(String[] args) {

        for (int i = 1; i < 51; i++) {

            if (isOdd(i)) {
                pi = pi+(4/((i*2)-1));
                // System.out.println(i);
            } else {
                pi = pi-(4/(i*2)-1);
                // System.out.println(i);
            }
        }

        System.out.println(pi);
    }

    private static boolean isOdd(int n) {
        if (n % 2 == 1) {
            return true;
        } else {
            return false;
        }
    }
}

FYI: The output for this is 28.

Thanks.

@Mr.M You're right, I'll try a different type. Thanks for the reminder :) — Waitdev_
– Waitdev_, Commented Mar 18, 2016 at 9:06
I didn't know that technique, but it's a slow one! The second decimal is guaranteed only from the 1256 iteration... — Olivier Grégoire
– Olivier Grégoire, Commented Mar 18, 2016 at 9:29

sfThomas · Accepted Answer · 2016-03-18 09:11:14Z

6

You are missing a parenthesis in your else block, and you need to force float calculation by e.g writing 4.0 instead of 4. It should read:

if (isOdd(i)) {
    pi = pi+(4.0/((i*2)-1));
    // System.out.println(i);
} else {
    pi = pi-(4.0/((i*2)-1));
    // System.out.println(i);
}

You also need to use a float or a double for pi.

Extra - as for your isOdd method, it could be simplified like:

private static boolean isOdd(int n) {
     return (n % 2 == 1)
}

But this does not influence the result.

edited Mar 18, 2016 at 9:11

answered Mar 18, 2016 at 9:03

sfThomas

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4 Comments

Waitdev_ Over a year ago

I tried this, and the result is 3.0 now. It's an improvement, but not what I was hoping. Thanks though.

Madhawa Priyashantha Over a year ago

@Waitdev_ 4/((i*2) still give you int value.because int/int=int

sfThomas Over a year ago

You might want to write 4.0 instead of 4 - that'll force the usage of non-integers. Uptaded the answer.

diginoise Over a year ago

Since using Gregory-Leibniz series could lead to millions of iterations, odness can be established a tad faster by using (i & 0x01) > 0

Roel Strolenberg · Accepted Answer · 2016-03-18 09:09:21Z

0

First off: an int is an integer, whole number. you want to change it to:

static double pi = 0;

Otherwise the answer always gets rounded down to a whole number.

Secondly, you are missing parenthesis in your else statement:

} else {
    pi = pi-(4/((i*2)-1));
}

If you want to be completely safe, change your integers in the if and else clause to doubles:

            if (isOdd(i)) {
                pi = pi+(4.0/((i*2.0)-1.0));
                // System.out.println(i);
            } else {
                pi = pi-(4.0/((i*2.0)-1.0));
                // System.out.println(i);
            }

But that shouldn't be neccessary

edited Mar 18, 2016 at 9:09

answered Mar 18, 2016 at 9:07

Roel Strolenberg

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1 Comment

Mark Over a year ago

yes it should. otherwise i = 0-2 will always be pi +/- 1 and 3+ will always be pi +/- 0. Though he only needs one .0, not all of them I guess.

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